Question

$$\begin{aligned}&\text { If } w=f(x, y) \text { and } x=r \cos \theta, y=r \sin \theta, \text { find formulas for }\\\ &\partial w / \partial r, \partial w / \partial \theta, \text { and } \partial^{2} w / \partial r^{2}\end{aligned}$$.

Short answer

1) \( \frac{\partial w}{\partial r} = \frac{\partial w}{\partial x} \cos \theta + \frac{\partial w}{\partial y} \sin \theta\); 2) \( \frac{\partial w}{\partial \theta} = -r \sin \theta \frac{\partial w}{\partial x} + r \cos \theta \frac{\partial w}{\partial y}\); 3) \(\frac{\partial^2 w}{\partial r^2} = \cos \theta \frac{d}{dr} (\frac{\partial w}{\partial x}) + \sin \theta \frac{d}{dr} (\frac{\partial w}{\partial y}) \)

Step-by-step solution

Understand the Problem

We need to find the partial derivatives of the function \(w=f(x, y)\) concerning polar coordinates \(r\) and \(\theta\). Specifically, we need formulas for: \(\frac{\partial w}{\partial r}, \frac{\partial w}{\partial \theta},\text{ and }\frac{\partial^2 w}{\partial r^2}\).

Use the Chain Rule for Partial Derivatives

Given that \(x=r \cos \theta\) and \(y=r \sin \theta\), we use the chain rule. The partial derivatives are given by:\[\frac{\partial w}{\partial r} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial r}\]\[\frac{\partial w}{\partial \theta} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial \theta}\]

Compute Derivatives of Coordinates

Calculate the partial derivatives of the coordinates:\[\frac{\partial x}{\partial r} = \cos \theta \quad \text{and} \quad \frac{\partial y}{\partial r} = \sin \theta\]\[\frac{\partial x}{\partial \theta} = -r \sin \theta \quad \text{and} \quad \frac{\partial y}{\partial \theta} = r \cos \theta\]

Find \(\frac{\partial w}{\partial r}\)

Substitute the derivatives into the chain rule for \(\frac{\partial w}{\partial r}\) and simplify:\[\frac{\partial w}{\partial r} = \frac{\partial w}{\partial x} \cos \theta + \frac{\partial w}{\partial y} \sin \theta\]

Find \(\frac{\partial w}{\partial \theta}\)

Substitute the derivatives into the chain rule for \(\frac{\partial w}{\partial \theta}\) and simplify:\[\frac{\partial w}{\partial \theta} = \frac{\partial w}{\partial x} (-r \sin \theta) + \frac{\partial w}{\partial y} (r \cos \theta)\]\[\frac{\partial w}{\partial \theta} = -r \sin \theta \frac{\partial w}{\partial x} + r \cos \theta \frac{\partial w}{\partial y}\]

Find \(\frac{\partial^2 w}{\partial r^2}\)

Differentiate \(\frac{\partial w}{\partial r}\) with respect to \(r\) again:\[\frac{\partial^2 w}{\partial r^2} = \frac{\partial}{\partial r} \left( \frac{\partial w}{\partial x} \cos \theta + \frac{\partial w}{\partial y} \sin \theta \right)\]By applying the chain rule again for each term separately:\[\frac{\partial^2 w}{\partial r^2} = \cos \theta \frac{\partial}{\partial r} \left( \frac{\partial w}{\partial x} \right) + \sin \theta \frac{\partial}{\partial r} \left( \frac{\partial w}{\partial y} \right)\]These are further expanded by again using the previously defined relations between partial derivatives.

Key concepts

chain rule

One of the main tools to solve partial derivatives in polar coordinates is the chain rule. The chain rule in multivariable calculus allows us to find the derivative of a composite function. Here, we have a function \( w = f(x, y) \), and we know that the variables \( x \) and \( y \) themselves depend on \( r \) and \( \theta \). To find the partial derivatives of \( w \) with respect to \( r \) and \( \theta \), we express how \( x \) and \( y \) change with respect to \( r \) and \( \theta \). This is where the chain rule comes into play.
For example, to find \( \frac{\frac{\text{partial}}{\frac{\text{partial}r}} \), we use the chain rule:
\[\frac{\text{partial}}{\text{partial}r} = \frac{\text{partial}}{\text{partial}x} \frac{\text{partial}x}{\text{partial}r} + \frac{\text{partial}}{\text{visional}y} \frac{\text{partial}y}{\text{partial}r}\]
By calculating the partial derivatives of \( x \) and \( y \) with respect to \( r \), we can substitute those into the equation. This approach is used to find the formulas for \( \frac{\text{partial}w}{\text{partial}r} \), \( \frac{\text{partial}w}{\text{partial}\theta} \), and \( \frac{\text{partial}^{2}w}{\text{partial}r^{2}} \).

polar coordinates

In polar coordinates, the position of a point in the plane is defined by its distance from the origin \( r \) and the angle \( \theta \) it makes with the positive x-axis. This contrasts with Cartesian coordinates where we define a point by \( x \) and \( y \) coordinates. In our problem, we used the following transformations:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

These transformations allow us to convert a problem from Cartesian coordinates to polar coordinates.

partial derivatives

Partial derivatives allow us to measure how a function changes as one of its variables changes, keeping other variables constant. In our problem, \( w = f(x, y) \) is given in terms of two variables, \( x \) and \( y \). To find the partial derivatives of \( w \) with respect to \( r \) and \( \theta \), we use the chain rule to combine their respective partial derivatives.
For instance, to compute \( \frac{\text{partial}w}{\text{partial}r} \), we determined:
\[ \frac{\text{partial}w}{\text{partial}r} = \frac{\text{partial}w}{\text{partial}x} \cos\theta + \frac{\text{partial}w}{\text{partial}y} \sin\theta \] Using similar steps, we find other partial derivatives. These derivatives are crucial for understanding how the function behaves in different directions around a particular point.

multivariable calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of multiple variables. In our problem, the function \( w = f(x, y) \) involves two independent variables, \( x \) and \( y \), which depend on other variables \( r \) and \( \theta \). Through multivariable calculus, we study derivatives and integrals of these functions.
Important concepts in multivariable calculus related to this problem include:

• The chain rule, which helps us find partial derivatives of composite functions
• Coordinate transformations, like the conversion between Cartesian and polar coordinates
• Second-order partial derivatives, important for understanding curvature and other properties of surfaces

Multivariable calculus provides the tools to analyze more complex problems, which is essential in fields like physics, engineering, and economics.